A Way of Learning Mathematics

A Way of Learning Mathematics

  M.Velrajan

Learning takes place when the learner has freedom to think, questions arise into the learner and the learner tries to get answers for that. One can also learn by observation and from any incidences. Thinking beyond, Thinking differently and asking questions themselves and to others are essential to develop learning.

Learning by observation

We illustrate a way of learning by simple observation on numbers. 

Look into the multiplication table of 9.

  1 ⨯ 9 = 09 

  2 ⨯ 9 = 18

  3 ⨯ 9 = 27

  4 ⨯ 9 = 36

  5 ⨯ 9 = 45

  6 ⨯ 9 = 54

  7 ⨯ 9 = 63

  8 ⨯ 9 = 72

  9 ⨯ 9 = 81

10 ⨯ 9 = 90

What do we observe in the multiplication table of 9 ?

Observation :

Sum of the digits of all in the RHS is 9, the first (tenth place) digits in the RHS are 0, 1, 2, . . . , 9 and the second (unit place) digits are 9, 8, 7, . . . , 1, 0 and each row has a pattern, for example,

          6 ⨯ 9 =  5 4

          5 = 6 - 1, 4 = 9 - 5 or 4 = 10 - 6.

Why ?

6 ⨯ 9 = 6 ⨯ (10 -1) = (5 + 1)(10 - 1) = 5 ⨯ 10 - 5 + (10 - 1) = 5 ⨯ 10 + (9 - 5) = 5 ⨯ 10 + (10 - 6)    

6 ⨯ 9 = 5 ⨯ 10 + (9 - 5) =  5 4

Thinking Beyond :

9 = 10 - 1

99 = 100 - 1 = 102 - 1

999 = 1000 - 1 = 103 - 1

…………………..

Can we extend ?

What is 53 ⨯ 99 = _ _  _ _

Yes

53 ⨯ 99 = 52 47

52 = 53 - 1 and 47 = 99 - 52 or 4 = 9 - 5, 2 = 9 - 7

   53 ⨯ 99 = 52 ⨯ 100 + (99 - 52) = 52 ⨯ 100 + (100 - 53)

     7 ⨯ 99 =   06 93 = 693

279 ⨯ 999 = 278 721

  79 ⨯ 999 = 078 921

Why ?

In general,

a(10n - 1) = (a - 1) (10n - 1) + (10n - 1)

               = (a - 1) 10n – (a - 1) + (10n - 1)

            = (a - 1) 10n + [(10n - 1) – (a - 1) ]

               = (a - 1) 10n + (10n  – a)

Also, if we write table for 99, from 1 ⨯ 99 to 100 ⨯ 99, sum of the digits of all in the RHS is 9, the first two (1000th and 100th place) digits in the RHS are 00, 01, 02, . . . , 98, 99 and the last two (10th and unit place) digits are 99, 98, 97, . . . , 01, 00.  

This is a way of learning. i.e gaining knowledge from the 9th table, understanding it thoroughly, applying the knowledge properly and creating.

And it is also the way of learning as per Thiruvalluvar in the chapter Education in the Thirukkural :  

"கற்க கசடற கற்பவை கற்றபின் 

நிற்க அதற்குத் தக"

Read as :  "Karka kasadara karpavai katrapin 

                  Nirka atharku thaga” 

Meaning : “Learn thoroughly whatever to be learnt, after learning  

                   live according it"

Observe multiplication by 11.

08 ⨯ 11 = 088     0 8 0 8     8 + 0 = 8

10 ⨯ 11 = 110     1 0 1 0     0 + 1 = 1

11 ⨯ 11 = 121     1 1 1 1     1 + 1 = 2

. . . . . . . . . ..

18 ⨯ 11 = 198     1 8 1 8     8 + 1 = 9

19 ⨯ 11 = 209     1 9 1 9     9 + 1 = 10     1 10 9    1 + 1 = 2

38 ⨯ 11 = 418     3 8 3 8     8 + 3 = 11     3 11 8    3 + 1 = 4

99 ⨯ 11 = 1089   9 9 9 9     9 + 9 = 18     9 18 9   9 + 1 = 10

112 = 121

1112 = 12321

11112 = 1234321

. . . . . .

11…12 = 123….(n-1)n(n-1)....21

   n 1s

332 = 9 ⨯ (121) = 1089

3332 = 9 ⨯ (12321) = 110889

33332 = 9 ⨯ (1234321) = 11108889

333332 = 9 ⨯ (123454321) = 1111088889

. . . . . . . .  

333…32 = 9 ⨯ (123….(n-1)n(n-1)....21) = 11….1110888….8889

   n 3s                                                          (n-1) 1s, 0, ( n-1) 8s, 9

772 = 49 ⨯ (121) = (50-1) ⨯ (121) = 6050 - 121 = 5929

112 = 121

1012 = 10201

10012 = 1002001

100012 = 100020001

. . . . . . . .  .

What about division by 9 ?

1/9 = 0.111111………

2/9 = 0.222222………

3/9 = 0.333333………

4/9 = 0.444444………

5/9 = 0.555555………

6/9 = 0.666666………

7/9 = 0.777777………

8/9 = 0.888888………

Why ?

Let 0 < a < 10 and x = 0.aaaaa…..  . Then 10x = a.aaaaaa…. = a + x. Hence x = a/9. 

Hence 9/9 = 0.999999………    = 1. (Note that the decimal representation of 1 is not unique.)

Extension :

23/99 = 0.23232323…………

87/99 = 0.87878787………….

7/99  = 0.07070707………….

895/999 = 0.895895895…….

84/999 = 0.084084084………

8/999 = 0.008008008………

Think

136/333 = ?                28/33 = ?        7/33 = ?

103/111 = ?                 9/11 = ?

In 1873 Cantor proved the set of all rational numbers ℚ is countably infinite, i.e. there is a one-one correspondence between ℚ and ℕ the set of all natural numbers.

Theorem (Cantor)   Card (ℚ+) = Card (ℕ).

First we list all of ℚ+ (the positive rational numbers) in the following way

 

Now, as in the above drawing, weave our way through the positive rational numbers assigning the next natural number to a rational number we have not seen before. So, using the drawing above, we have the following correspondence between ℕ and ℚ+ :  

1 → 1/1,    2 → 1/2,    3 → 2/1,    4 → 3/1,   5 → 1/3 (skip over 2/2 since we have seen 2/2 = 1/1 before),   6 → 1/4,    7 → 2/3,   8 → 3/2,  . . . .

This correspondence is a one-to-one correspondence between ℕ and ℚ+. 

Thus card(ℕ) = card(ℚ+) and hence is countable. 

The argument given in the Proof is called the Cantor diagonalization argument.

Observe the uniqueness and the crux of the proof and think of where similar proof can be given.

Similarly, by including each negative rational immediately after the corresponding positive rational in the above list and placing 0 as the 0th row, we can show that card(ℕ) = card(ℚ).

Also we can show, in a similar way, that any countable union of countable sets is countable. 

( We can list the countable union of countable sets as

Set 1:  a11   a12    a13     …….. a1n   ……

Set 2:  a21   a22    a23     …….. a2n   ……

Set 3:  a31   a32    a33     …….. a3n   ……

.

.

.

Set n:  an1   an2    an3     …….. ann   ……

.

.

.

and use Cantor's diagonalization argument.)

So, if we learn the proof of Cantor's theorem properly, we can easily prove the above.

Hope this gives an idea of learning.